# Does the Cantor-Schröder-Bernstein Theorem hold in the category opposite to the category of noetherian commutative rings?

Let $$A$$ and $$B$$ be noetherian commutative rings with one, and let $$f:A\to B$$ and $$g:B\to A$$ be epimorphisms.

Are the rings $$A$$ and $$B$$ necessarily isomorphic?

[In this post "ring" means "commutative ring with one", and morphisms are required to map $$1$$ to $$1$$. By definition, a morphism of rings $$f:A\to B$$ is an epimorphism if for all pairs of morphisms $$(g,h):B\rightrightarrows C$$ the equality $$g\circ f=h\circ f$$ implies $$g=h$$. Surjective morphisms are epimorphic, but the converse does not always hold: for instance the inclusion $$\mathbb Z\to\mathbb Q$$ is an epimorphism. For more details about epimorphisms see this post.]

The busy reader is invited to skip the sequel.

The answers to the following variants of the above question are known:

(1) If $$f:A\to B$$ and $$g:B\to A$$ are injective morphisms of noetherian rings, are $$A$$ and $$B$$ necessarily isomorphic? The answer is No, as shown by the following example taken from a comment of Sam Lichtenstein to this interesting MathOverflow question. Let $$K$$ be a field, $$x$$ an indeterminate, $$f:K[x^2,x^3]\to K[x]$$ the inclusion, and $$g:K[x]\to K[x^2,x^3]$$ the (clearly injective) morphism defined by $$g(p(x))=p(x^2)$$. Note that $$K[x^2,x^3]$$ is not isomorphic to $$K[x]$$ because the ideal $$(x^2,x^3)$$ of $$K[x^2,x^3]$$ is not principal.

(2) If $$f:A\to B$$ and $$g:B\to A$$ are surjective morphisms of rings, are $$A$$ and $$B$$ necessarily isomorphic? The answer is No, as shown by the following example taken from the same comment of Sam Lichtenstein. Set $$A:=\mathbb Z/(4)\times\mathbb Z/(4)\times\cdots,\quad B:=\mathbb Z/(2)\times A,$$ let $$f:A\to B$$ be defined by $$f(x_1,x_2,\dots)=(h(x_1),x_2,\dots)$$, where $$h$$ is the unique ring morphism from $$\mathbb Z/(4)$$ to $$\mathbb Z/(2)$$, and let $$g:B\to A$$ be defined by $$g(x_1,x_2,\dots)=(x_2,x_3,\dots)$$. The rings $$A$$ and $$B$$ are not isomorphic because the equations $$2x=0$$ and $$x^2=x$$ have no nonzero simultaneous solutions in $$A$$, and one such solution in $$B$$ (namely $$x=(1,0,\dots)$$).

(3) If $$f:A\to B$$ and $$g:B\to A$$ are surjective morphisms of noetherian rings, are $$A$$ and $$B$$ isomorphic? The answer is Yes, because surjective endomorphisms of noetherian rings are isomorphisms. But epimorphic endomorphisms of noetherian rings are not always isomorphisms: see this answer of Eric Wofsey.

Edit: Crossposted on MathOverflow https://mathoverflow.net/q/343418/461

No because you can take $$A = \mathbf{Z}[x, 1/(x - n); n\geq 0]$$ and $$B = \mathbf{Z}[x, 1/x, 1/(x - n); n \geq 2]$$ and the maps are $$B \to A$$ is the inclusion and $$A \to B$$ sends $$x$$ to $$x - 2$$. The reason $$A$$ is not isomorphic to $$B$$ is that the gaps between the missing points'' are different for $$A$$ and $$B$$. More precisely, any isomorphism $$A \to B$$ sends $$x$$ to something of the form $$(a x + b)/(cx + d)$$ with $$a, b, c, d \in \mathbf{Q}$$ and there is no such function which sends the set $$\{0, 2, 3, \ldots\} \cup \{\infty\}$$ bijectively to $$\{0, 1, 2, 3, \ldots\} \cup \{\infty\}$$.